#dy/dx of logx/√(x^2-a^2)

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serene canopy
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i have solved them and now stuck at logx/x/√(x^2-a^2)

flint capeBOT
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feral drift
serene canopy
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wait

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let me get my mom's phone to take picture

feral drift
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alright

serene canopy
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here

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i somehow have to get it in logx/√(x^2-a^2) form

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for quotient rule

feral drift
# serene canopy

$\frac{dy}{dx}=\frac{\frac{d logx}{dx}\cdot\sqrt{x^{2}-a^{2}} - logx \cdot \frac{d \sqrt{x^{2}-a^{2}}}{dx}}{x^{2}-a^{2}}$

feral drift
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uh wait

feral drift
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derivate of log x is not 1/x

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its $\frac{1}{x lna}$ where a is the base

warm wingBOT
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rev #tea4life

serene canopy
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i havent derivated it

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yet

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i kept it for the quoatient rule

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isnt logx just log base e and x

warm wingBOT
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rev #tea4life

serene canopy
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what happened to the x in x/√(x^2-a^2)

feral drift
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yk what, i can't do shit with this bot

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I'll send pics

serene canopy
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okay

feral drift
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tell me one thing

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is the base in the log 10 or e

serene canopy
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e

feral drift
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then it's natural log

serene canopy
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yea

feral drift
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i thought it was base 10

serene canopy
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1/x

feral drift
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so it's d/dx will be 1/x

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yes now i know

serene canopy
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yea

feral drift
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here btw the sign between the two fractions is -ve

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did u get it till here?

serene canopy
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i used chain rule at the denominator

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lemme try againn

feral drift
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you brought the denominator up?

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like $(\sqrt{x^{2}-a^{2}})^{-1}$

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this?

serene canopy
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no

warm wingBOT
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rev #tea4life

serene canopy
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used it at the bottom

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like u = logx

feral drift
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I'd just use quotient rule

serene canopy
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v=√(x^2-a^2)

serene canopy
feral drift
serene canopy
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the answer is weird too

feral drift
shrewd urchin
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Hello

feral drift
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simplifying what i get

serene canopy
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yea

feral drift
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it'll be the same

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try quotient rule

serene canopy
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okay

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lemme try

feral drift
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silly error mb

serene canopy
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np

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thanks

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i got the answer

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should i close it?

feral drift
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yeah sure

feral drift
serene canopy
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+close